Special Isosceles Trapezoid

Say the parallel sides of the trapezoid have lengths $a$ and $b$ and the non-parallel legs have length $l$ ; let its area be $A$ and perimeter $P$ .

The angle bisectors of a quadrilateral meet in a single point if and only if the quadrilateral is tangential (see the link for a proof of this) - this means we can construct an incircle in the trapezoid.

As a tangential quadrilateral, we get two nice properties; firstly, $a+b=2l=\frac{P}{2}$

so that $P=4l$ ; secondly $A=\frac{rP}{2}$

where $r$ is the inradius of the trapezoid. For the trapezoid to exist, we need $2r \le l$ . For it to satisfy $a \neq b$ , we need $2r<l$ .

Putting everything together, we have $P=4l$ and $A=2rl<l^2=\frac{P^2}{16}$

Both of these are positive integers, so we need $P^2>16$ . The smallest possible perimeter is therefore $P=5$ , and we can take $A=1$ to minimise their total; finally $l=\frac54$ so the answer is $\boxed9$ .

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Just a sidenote - the condition $A<\frac{P^2}{16}$ holds for any non-square quadrilateral, so we could get to that earlier.

Label the isosceles trapezoid as follows:

Let $x = \angle SPU = \angle UPT = \angle TQU = \angle UQR$ and $y = \angle PSU = \angle USV = \angle URV = \angle URQ$ .

As a trapezoid, $\angle TPS + \angle PSV = 180°$ , so $2x + 2y = 180°$ , which means $x + y = 90°$ . Then by the angle sum of $\triangle PUS$ , $\angle PUS = 180° - (x + y) = 180° - 90° = 90°$ . Similarly, $\angle QUR = 90°$ .

Let $a = PU = QU$ , $b = SU = RU$ , and $c = PS = QR$ . By the Pythagorean Theorem on $\triangle PUS$ and $\triangle QUR$ , $a^2 + b^2 = c^2$ .

$\triangle UVS \sim \triangle PUS$ by AA similarity, so that $\cfrac{SV}{SU} = \cfrac{SU}{SP}$ , or $\cfrac{SV}{b} = \cfrac{b}{c}$ , which solves to $SV = \cfrac{b^2}{c}$ and makes $SR = \cfrac{2b^2}{c}$ . Also from the similarity, $\cfrac{UV}{PU} = \cfrac{SU}{SP}$ , or $\cfrac{UV}{a} = \cfrac{b}{c}$ , which solves to $UV = \cfrac{ab}{c}$ .

Also, $\triangle PTU \sim \triangle PUS$ by AA similarity, so that $\cfrac{PT}{PU} = \cfrac{PU}{PS}$ , or $\cfrac{PT}{a} = \cfrac{a}{c}$ , which solves to $PT = \cfrac{a^2}{c}$ and makes $PQ = \cfrac{2a^2}{c}$ . Also from the similarity, $\cfrac{TU}{US} = \cfrac{SU}{SP}$ , or $\cfrac{TU}{b} = \cfrac{a}{c}$ , which solves to $TU = \cfrac{ab}{c}$ .

The perimeter of the trapezoid is then $P = PQ + QR + RS + SP = \cfrac{2a^2}{c} + c + \cfrac{2b^2}{c} + c$ , which after substituting $a^2 + b^2 = c^2$ simplifies to $P = 4c$ .

The area of the trapezoid is then $A = \cfrac{1}{2}(PQ + SR)TV = \cfrac{1}{2}\Bigg(\cfrac{2a^2}{c} + \cfrac{2b^2}{c}\Bigg) \cfrac{2ab}{c}$ , which after substituting $a^2 + b^2 = c^2$ simplifies to $A = 2ab$ .

From $A = 2ab$ , $b = \cfrac{A}{2a}$ , and from $P = 4c$ , $c = \cfrac{P}{4}$ , so that $a^2 + b^2 = c^2$ becomes $a^2 + \cfrac{A^2}{4a^2} = \cfrac{P^2}{16}$ , which solves to $a^2 = \frac{1}{32}(P^2 \pm \sqrt{P^4 - 256A^2})$ .

To have a real answer, the discriminant $P^4 - 256A^2 \geq 0$ , and to be a trapezoid (instead of a rectangle), $a \neq b$ so the discriminant $P^4 - 256A^2 > 0$ , which solves to $P^2 > 16A$ , whose smallest positive integer solutions are $A = 1$ and $P = 5$ .

Therefore, $c = \cfrac{P}{4} = \cfrac{5}{4}$ , so that $p = 5$ , $q = 4$ , and $p + q = \boxed{9}$ .

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Interestingly, the other lengths of the trapezoid also calculate to non-integer values of $PQ = \frac{1}{20}(25 - 3\sqrt{41})$ , $RS = \frac{1}{20}(25 + 3\sqrt{41})$ , $TV = \frac{4}{5}$ , $PU = QU = \frac{1}{8}(\sqrt{41} - 3)$ , and $SU = RU = \frac{1}{8}(\sqrt{41} + 3)$ .